


Polynomials
A polynomial is an expression that
contains one or more algebraic terms, each consisting of a constant
multiplied by a variable raised to a power greater than or equal
to zero. For example, is a polynomial with three terms
(the third term is . , on the other hand, is not a polynomial
because x is raised to a negative power. A binomial is
a polynomial with exactly two terms: and are both binomials.
The rest of this chapter will show you how to perform
different operations on and with polynomials.
Multiplying Binomials
There is a very simple acronym that is useful in remembering
how to multiply binomials. It is FOIL, and it stands
for First, Outer, Inner, Last.
This is the order that you multiply the terms of two binomials to
get the right product.
For example, if asked to multiply the binomials:
You first multiply the first terms of each binomial:
Next, multiply the outer terms of the binomials:
Then, multiply the inner terms:
Finally, multiply the last terms:
Combine like terms and you have your product:
Here are a few more examples:
Multiplying Polynomials
Every once in a while, the Math IC test will ask you to
multiply polynomials. It may seem like a daunting task. But when
the process is broken down, multiplying polynomials requires nothing
more than distribution and combining like terms.
Consider the polynomials (a + b + c)
and (d + e + f). To find their product, just distribute the
terms of the first polynomial into the second polynomial individually
and combine like terms to formulate your final answer:
Here’s another example:
As you can see, multiplying polynomials is little more
than rote multiplication and addition.
Quadratic Equations
A quadratic, or quadratic polynomial, is a polynomial
of the form ax^{2} + bx + c,
where a ≠ 0. The following polynomials are quadratics:
A quadratic equation sets a quadratic polynomial
equal to zero. That is, a quadratic equation is an equation of the
form ax^{2} + bx + c =
0. The values of x for which the equation holds are
called the roots, or solutions, of the quadratic equation.
Most of the questions on quadratic equations involve finding their
roots.
There are two basic ways to find roots: by factoring and
by using the quadratic formula. Factoring is faster,
but it can’t always be done. The quadratic formula takes longer
to work out, but it works for all quadratic equations. We’ll study
both in detail.
Factoring
To factor a quadratic, you must express it as the product
of two binomials. In essence, factoring a quadratic involves a reverseFOIL
process. Take a look at this quadratic:
In the example above, the leading term has a coefficient
of 1 (since 1x^{2} is the
same as x^{2}). Since the
two x variables are multiplied together during
the FIRST step of foiling to get the first term of the quadratic
polynomial, we know that the binomials whose product is this quadratic
must be of the form (x + m)(x + n),
where m and n are constants. You
also know that the sum of m and n is
10, since the 10x is derived from multiplying the
OUTER and INNER terms of the binomials and then adding the resulting
terms together (10x = mx + nx,
so m + n must equal 10). Finally,
you know that the product of m and n equals
21, since 21 is the product of the two last terms of the binomials.
Now you just need to put the pieces together to find the
values of m and n. You know that x is
the first term of both binomials, and you know that the sum of m and n is
10 and the product of m and n is
21. The pair of numbers that fit the bill for m and n are
3 and 7. Thus, x^{2} +
10x + 21 = (x + 3)(x +
7). The quadratic expression has now been factored and simplified.
On the Math IC, though, you will often be presented with
a quadratic equation. The only difference between
a quadratic equation and a quadratic expression is that the equation
is set equal to 0 (x^{2} +
10x + 21 = 0). If you have such an equation, then
once you have factored the quadratic you can solve it. Because the
product of two terms is zero, one of the terms must be equal to
zero. Thus, since x + 3 = 0 or x +
7 = 0, the solutions (also known as the roots) of the quadratic
must be x = –3 and x = –7.
Quadratics with Negative Terms
So far we’ve dealt only with quadratics in which
the terms are all positive. Factoring a quadratic that has negative
terms is no more difficult, but it might take slightly longer to
get the hang of it, simply because you are less used to thinking
about negative numbers.
Consider the quadratic equation x^{2} –
4x – 21 = 0. There are a number of things you can tell
from this equation: the first term of each binomial is x,
since the first term of the quadratic is x^{2};
the product of m and n is –21;
and the sum of a and b equals
–4. The equation also tells you that either m or n must
be negative but that both cannot be negative, because the multiplication
of one positive and one negative number can only result in a negative number.
Now you need to look for the numbers that fit these requirements
for m and n. The numbers that
multiply together to give you –21 are: –21 and 1, –7 and 3, 3 and
–7, and 21 and –1. The pair that works in the equation is –7 and
3.
Two Special Quadratic Polynomials
There are two special quadratic polynomials that pop up
quite frequently on the Math IC, and you should memorize them. They
are the perfect square and the difference of two squares. If you
memorize the formulas below, you may be able to avoid the time taken
by factoring.
There are two kinds of perfect square quadratics. They
are:
 a^{2} + 2ab + b^{2} = (a + b)(a + b) = (a + b)^{2}. Example: a^{2} + 6ab + 9 = (a + 3)^{2}
 a^{2} – 2ab + b^{2} = (a – b)(a – b) = (a – b)^{2}. Example: a^{2} – 6ab + 9 = (a –3)^{2}
Note that when you solve for the roots of a perfect
square quadratic equation, the solution for the equation (a + b)^{2} =
0 will be –b, while the solution for (a + b)^{2} =
0 will be b.
The difference of two squares quadratics follow the form
below:
Here’s an instance where knowing the perfect square or
difference of two square equations can help you:

To solve this problem by working out the math, you would
do the following:
If you got to the step where you had 2(x^{2} +
10x +25) = 0 and realized that you were working
with a perfect square of 2(x + 5)^{2},
you could immediately have divided out the 2 from both sides of
the equation and seen that the solution to the problem is –5.
Practice Quadratics
Since the ability to factor quadratics relies in large
part on your ability to “read” the information in the quadratic,
the best way to get good is to practice, practice, practice. Just
like perfecting a jump shot, repeating the same drill over and over
again will make you faster and more accurate. Take a look at the
following examples and try to factor them on your own before you
peek at the answers.
The Quadratic Formula
Factoring using the reverseFOIL method is really only
practical when the roots are integers. Quadratics, however, can
have decimal numbers or fractions as roots. Equations like these
can be solved using the quadratic formula. For an equation of the
form ax^{2} + bx + c = 0,
the quadratic formula states:
Consider the quadratic equation x^{2} +
5x + 3 = 0. There are no integers with a sum of
5 and product of 3. So, this quadratic can’t be factored, and we
must resort to the quadratic equation. We plug the values, a =
1, b = 5, and c = 3 into the formula:
The roots of the quadratic are approximately {–4.303,
–.697}.
Finding the Discriminant:
If you want to find out quickly how many roots an equation
has without calculating the entire formula, all you need to find
is an equation’s discriminant. The discriminant of
a quadratic is the quantity b^{2} –
4ac. As you can see, this is the radicand in the
quadratic equation. If:
 b^{2} – 4ac = 0, the quadratic has one real root and is a perfect square.
 b^{2} – 4ac > 0, the quadratic has two real roots.
 b^{2} – 4ac < 0, the quadratic has no real roots, and two complex roots.
This information is useful when deciding whether to crank
out the quadratic formula on an equation, and it can spare you some
unnecessary computation. For example, say you’re trying to solve
for the speed of a train in a rate problem, and you find that the
discriminant is less than zero. This means that there are no real
roots (a train can only travel at speeds that are real numbers),
and there is no reason to carry out the quadratic formula.
